Mk giải 1 câu thôi dc hok bn ?
a) Ta có: \(\left(n+3\right)⋮\left(n+1\right)\)
\(\Leftrightarrow\left(n+1+2\right)⋮\left(n+1\right)\)
Vì \(\left(n+1\right)⋮\left(n+1\right)\forall n\in N\)
nên \(2⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(2\right)\)
\(\Leftrightarrow n+1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow n\in\left\{0;-2;1;-3\right\}\)
mà \(n\in N\)
nên \(n\in\left\{0;1\right\}\)
Vậy: \(n\in\left\{0;1\right\}\)
b) Ta có: \(\left(2n+6\right)⋮\left(2n-6\right)\)
\(\Leftrightarrow\left(2n-6+12\right)⋮\left(2n-6\right)\)
Vì \(\left(2n-6\right)⋮\left(2n-6\right)\forall n\in N\)
nên \(12⋮2n-6\)
\(\Leftrightarrow2n-6\inƯ\left(12\right)\)
\(\Leftrightarrow2n-6\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
\(\Leftrightarrow2n\in\left\{7;5;8;4;9;3;10;2;12;0;18;-6\right\}\)
\(\Leftrightarrow n\in\left\{\frac{7}{2};\frac{5}{2};4;2;\frac{9}{2};\frac{3}{2};5;1;6;0;9;-3\right\}\)
Vì \(n\in N\)
nên \(n\in\left\{4;2;5;1;6;0;9\right\}\)
Vậy: \(n\in\left\{4;2;5;1;6;0;9\right\}\)
c) Ta có: \(\left(2n+3\right)⋮\left(n-2\right)\)
\(\Leftrightarrow\left(2n-4+7\right)⋮\left(n-2\right)\)
Vì \(\left(2n-4\right)⋮\left(n-2\right)\forall n\in N\)
nên \(7⋮\left(n-2\right)\)
\(\Leftrightarrow n-2\inƯ\left(7\right)\)
\(\Leftrightarrow n-2\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow n\in\left\{3;1;9;-5\right\}\)
Vì \(n\in N\)
nên \(n\in\left\{3;1;9\right\}\)
Vậy: \(n\in\left\{3;1;9\right\}\)
d) Ta có: \(\left(3n+2\right)⋮\left(n-3\right)\)
\(\Leftrightarrow\left(3n-9+11\right)⋮\left(n-3\right)\)
Vì \(\left(3n-9\right)⋮\left(n-3\right)\)
nên \(11⋮\left(n-3\right)\)
\(\Leftrightarrow n-3\inƯ\left(11\right)\)
\(\Leftrightarrow n-3\in\left\{1;11;-1;-11\right\}\)
\(\Leftrightarrow n\in\left\{4;14;2;-8\right\}\)
Vì \(n\in N\)
nên \(n\in\left\{4;14;2\right\}\)
Vậy: \(n\in\left\{4;14;2\right\}\)
a) ( n + 3 ) ⋮ ( n + 1 )
Do đó ta có ( n + 3 ) = ( n + 1 ) + 2
Nên 2 ⋮ n + 1
Vậy n + 1 ∈ Ư(2) = {-1; 1; -2; 2}
Ta có bảng sau :
| n + 1 | -1 | 1 | -2 | 2 |
| n | -2 | 0 | -3 | 1 |
➤ Vậy n ∈ {-2; 0; -3; 1}
b) ( 2n + 6 ) ⋮ ( 2n - 6 )
Do đó ta có ( 2n + 6 ) = ( 2n - 6 ) + 12
Nên 12 ⋮ 2n - 6
Vậy 2n - 6 ∈ Ư(12) = {-1; 1; -2; 2; -3; 3; -4; 4; -6; 6; -12; 12}
Ta có bảng sau :
| 2n - 6 | -1 | 1 | -2 | 2 | -3 | 3 | -4 | 4 | -6 | 6 | -12 | 12 |
| 2n | 5 | 7 | 4 | 8 | 3 | 9 | 2 | 10 | 0 | 12 | -6 | 18 |
| n | 2,5 | 3,5 | 2 | 4 | 1,5 | 4,5 | 1 | 5 | 0 | 6 | -3 | 9 |
Vì n ∈ Z nên ta loại bỏ các số thập phân như : 2,5 ; 3,5 ; 1,5 ; 4,5
➤ Vậy n ∈ {2; 4; 1; 5; 0; 6; -3; 9}
c) ( 2n + 3 ) ⋮ ( n - 2 )
⇒ \(\left[{}\begin{matrix}\text{( 2n + 3 ) ⋮ ( n - 2 )}\\\text{( n - 2 ) ⋮ ( n - 2 )}\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}\text{( 2n + 3 ) ⋮ ( n - 2 )}\\\text{2( n - 2 ) ⋮ ( n - 2 )}\end{matrix}\right.\)
Do đó ta có ( 2n + 3 ) ⋮ 2( n - 2 )
Mà ( 2n + 3 ) = 2( n - 2 ) + 7
Nên 7 ⋮ n - 2
Vậy n - 2 ∈ Ư(7) = {-1; 1; -7; 7}
Ta có bảng sau :
| n - 2 | -1 | 1 | -7 | 7 |
| n | 1 | 3 | -5 | 9 |
➤ Vậy n ∈ {1; 3; -5; 9}
d) ( 3n + 2 ) ⋮ ( n - 3 )
⇒ \(\left[{}\begin{matrix}\text{( 3n + 2 ) ⋮ ( n - 3 )}\\\text{( n - 3 ) ⋮ ( n - 3 )}\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}\text{( 3n + 2 ) ⋮ ( n - 3 )}\\\text{3( n - 3 ) ⋮ ( n - 3 )}\end{matrix}\right.\)
Do đó ta có ( 3n + 2 ) ⋮ 3( n - 3 )
Mà ( 3n + 2 ) = 3( n - 3 ) + 11
Nên 11 ⋮ n - 3
Vậy n - 3 ∈ Ư(11) = {-1; 1; -11; 11}
Ta có bảng sau :
| n - 3 | -1 | 1 | -11 | 11 |
| n | 2 | 4 | -8 | 14 |
➤ Vậy n ∈ {2; 4; -8; 14}
